you begin by treating the inequality just as you would an equal sign. what would you do if you had (x + 3)(x - 3) = 0 ? well, if two numbers being multiplied gives zero, then one or the other must be zero, so we would split it up into two equations: x + 3 = 0 and x - 3 = 0. A similar thing is to be done here

(x + 3)(x - 3) > 0

=> x + 3 > 0 or x - 3 > 0

=> x > -3 or x > 3

now, with inequalities, things sometimes work out weird, and in evaluating algebraically, you might get the signs turned the wrong way. so here we test the answers. we have two values to work around, x = -3 and x = 3. x cannot be equal to either of these values since that would give zero, and we must be greater than zero. so now, let's check to the left and right of these numbers.

if x < - 3, is (x + 3)(x - 3) > 0 ? .........YES! so this is one solution

We need to have a little background knowledge here, you should know that:

Now let's see how to do this:

.............applied the identity i mentioned above

..............got everything to one side

or

Now, what values of satisfy these equations, such that ?

5. (x+1)^2(x-2) + (x+1)(x-2)^2=0

Now this looks atrotous, but it's actually not that bad. with a keen eye you will realize that there are common terms here, they don't look like regular terms, like say, x, but they are groups. we have (x + 1) and (x - 2) being common to both terms here, so let's factor them out and see where it gets us

...........factor out the common terms

.........simplify what's in the square brackets

Can you take it from here?

6. 27^2x=9^x-3

Please use parentheses. it seems that (x - 3) is the power of 9, but i'm not sure. be careful when you are typing math, brackets are important tools to clarify what you mean

i'll assume x - 3 is the power of 9 (which means you should have typed 27^(2x) = 9^(x - 3) )

I will further assume that you are familiar with the laws of exponents, if not, look them up.

this is a nice exponential equation, we can express both side in terms of the same base. Note that 27 = 3^3 and 9 = 3^2, so we can replace 27 and 9 with their representations in base 3

Now we multiply the powers (i hope you know why, here's where your knowledge of the laws of exponents comes into play)

Now if the bases are the same, the powers have to be the same to maintain equality, so we can equate the powers.

the rest is trivial, i leave it to you

7. logx + log(x-3)=1

Ah, logarithms. I hope you know the rules. here are the ones we need for this problem:

Law 1:
If , then

Law 2:

Now, on to the problem (I assume we are using log to the base 10, as is reasonable):